Dimensional analysis of the Lagrangian

In his 1995 paper, Kaplan explains what are relevant, irrelevant and marginal interactions. The idea is this: the action, S, has dimension $\hbar$. When taking $\hbar=c=1$, $[S]=0$. Besides that $[x]=-1$. so from $$S=\int d^4x {\cal L}$$ we conclude $$[{\cal L}]=4.$$ This means each of the terms of ${\cal L}$ will have dimension 4. Say in a mass term of a scalar field $$\frac{1}{2} m^2\phi^2$$ $[m]=1$ because $[\phi]=1$, which makes the term to have dimension 4.

He says if the coefficients have negative dimension (different from our example) then the cross-section or decay width becomes smaller as the energy of the interaction $E$ becomes smaller, thus we call these interactions irrelevant. My question is why $[\rm coeff.]<0$ imply smaller cross-sections and decays widths as $E$ decreases?

• Actually, the dimension of $\phi$ follows from the dimension of $m$, not the other way round. The energy dimension of $m=E/c^2$ is 1 by definition. – J.G. Nov 19 '16 at 23:22

The scattering amplitudes are adimensional. The contribution from a Feynman diagram with $n$ ocurrences of a vertex is proportional to $g^n$ where $g$ is the corresponding coupling contant. The only other dimensionful quantity in the diagram is the energy $E$ of the particles involved in the interaction. If $[g]=-m<0$ then the diagram must be proportional to $E^{(m\cdot n)}g^n$, so cross sections decrease when the energy does.
If the coefficient has a negative dimension, then it would act like an inverse mass. Say for instance we have a coefficient $$\sim \frac{1}{m^2} ,$$ then the scattering amplitude at a given energy $E$ (assuming $m$ and $E$ are the only scale parameters) would go like $${\cal M} \sim \frac{E^2}{m^2} .$$ The value of $m$ is fixed, but the value of $E$ decreases with decreasing energy. Hence, the scattering amplitude would decrease at lower energies.